For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine variety X=Spec R and the sheaf of modules associated to M. What is the projective resolution doing geometrically to this sheaf?

Projectives are locally free sheaves, so if M itself is not projective then it must have some sort of "sharp twisting" or "pinching". In some way a projective resolution is "un-pinching" M. Geometrically, is this the same "un-pinching" that happens in a resolution of a singularity of a variety? Is there an example in low dimensions where one can actually draw a picture of this happening for modules?

One thing to think about, and in the direction of Ilya Nikokoshev's answer below, is when R is the algebra of continuous functions on a topological space (with some more conditions, but this is a comment, not an answer). Then projective modules are the same as vector bundles, and there are lots of geometric reasons why vector bundles are easier to deal with than general (sheaves of) modules.
–
Theo Johnson-FreydOct 19 '10 at 2:27

7 Answers
7

if M itself is not projective then it must have some sort of "sharp twisting" or "pinching".

This isn't quite right. You're conflating "projective/free" with "smooth". For example, M might be a quotient ring R/J, corresponding to a subvariety of X = Spec R, which subvariety might well be smooth, though R/J will never be projective (except in boring cases).

In fact, from this point of view the modules in the free resolution can't have anything to do with the geometry of M, since those free modules are just copies of the whole space. They can't carry any geometric information at all. The maps in the free resolution, on the other hand, carry all kinds of geometric information (Hilbert function, regularity, etc.).

The overuse of the word 'resolution' might be to blame here. Free (resp injective, flat, flasque,...) resolutions really have nothing to do with resolutions of singularities.

In my mind, the easiest way for a sheaf to not be locally free is for some point p to lack a local trivialization. For vector bundles I can visualize this happening when the stalks do not vary nicely: at p they change directions suddenly (what I meant by "sharply twisting") or the dimension of the stalks change ("pinching" of some sort). This might be confusing smoothness with local free-ness; I'm not really sure what's going on geometrically - hence my question. I feel like the maps in the resolution can be described geometrically rather than algebraically in some way. I just don't know how.
–
Justin DeVriesOct 23 '09 at 20:21

@Graham, you're right -- I misunderstood the question and your answer and thus replied in a very obnoxious manner. Sorry, let this comment remain here, but I deleted the irrelevant ones before.
–
Ilya NikokoshevNov 18 '09 at 20:51

I'd like to build on Graham and anonymous' answers. There are lots of ways that a free resolution can be used to obtain interesting geometric information, but I think of it in a rather different way than the "untwisting" suggested in your question. This is most easy to discuss in the case that you are studying a projective scheme $X\subseteq \mathbb P^n$. Let $S_X$ be the homogeneous coordinate ring $S_X$ of $X$, considered as a graded $k[x_0, \dots, x_n]$-module. Then, you can define the minimal free resolution of $S_X$:

$
0\to F_{n+1}\to \dots \to F_1\to F_0\to S_X \to 0
$

where each $F_i$ is a free, graded module over the polynomial ring.

As Graham points out, the maps in the free resolution carry a ton of information. But since we are in the graded case, even if you forget about the maps, you still obtain a lot of information. For instance, if you only know the graded ranks of each $F_i$, then you can recover the Hilbert function of $X$, as well as some other interesting numerical invariants: Hilbert polynomial, Castelnuovo-Mumford regularity, depth. You can also determine if $S_X$ is a Cohen-Macaulay or Gorenstein ring, simply from the graded ranks of the $F_i$.

Those are all straightforward applications: but there is also a large body of literature on how to extract more subtle geometric information about $X$ from the ranks of the $F_i$. This is essentially the focus of Eisenbud's "The Geometry of Syzygies," and that book is full of surprising examples.

I think that one example can be quite enlightening, so I'll yank Theorem 2.8 from Eisenbud's book. This says that if $X$ is the union of $7$ points in linearly general position in $\mathbb P^3$, then these $7$ points lie on a twisted cubic curve if and only if $\text{rank}(F_1)>4$. In other words: in this case, the rank of $F_1$ measures a subtle geometric property of the $7$ points (i.e. whether or not they lie on a twisted cubic).

It might be worth noting that projective $\mathcal O_X$-modules rarely exist as soon as you leave the affine world. For instance $\mathcal O_{\mathbb P^1_{\mathbb C}}$ is not projective or even the surjective image of a projective.

But you asked about an affine scheme, so we are OK, but perhaps then we can just restrict to free resolutions.

Perhaps the most obvious reason responsible for a module not being free is torsion
and perhaps one can have a geometric intuition about those.

The geometric intuition about torsion is that it is supported on a proper (meaning not the entire) subscheme, kind of like the structure sheaf of a subscheme, or several copies $\oplus$-ed together, or doing this for various subschemes and combining them. However, resolutions are about replacing sections by other ones, and in the case of torsions, lifting them to the ambient scheme. It might be only possible to do locally, but that's why we have sheaves. For example for the structure sheaf of a subscheme the natural first step is to lift the sections to the structure sheaf and continue with the ideal sheaf.

Ideal sheaves are typically not free, but for different reasons. If the ambient scheme is reduced and irreducible, then ideal sheaves are torsion-free, but that does not mean free in general.
For example the ideal sheaf of a subscheme of codimension at least $2$ needs at least two generators along the subscheme it defines, but it is isomorphic to the structure sheaf everywhere else, so in particular its rank is 1. The most meaningful resolution of an ideal sheaf is basically to take as many free elements as the minimal number of generators needed for the point that needs the most. In other words the first step of the resolution of an ideal sheaf is about how many functions can define the corresponding subscheme. Then the next one is about the relations among these functions and so on and on.

I would say it is clear that the farther one goes in the resolution it must be harder and harder to give meaning to the next step. So, perhaps the best is to think about it recursively; every new syzygy is the first syzygy of the previous one. That we might have some vague understanding. And the good news is, as long as the projective dimension of our module is finite, these sheaves will be getting nicer and nicer until one of them will be free already.

One important application of projective/locally free resolution
is the computation of intersection numbers.
Let $X$ be a smooth projective variety over a field $k$, and $Z_1,Z_2$ subschemes of $X$
of complementary dimension. Let $L^*$ be a locally free resolution of $\mathcal{O}_{Z_1}$ then the intersection number $(Z_1 . Z_2)$ equals $\chi(X,L^* \otimes \mathcal{O}_{Z_2})$ (Serre's formula).

While the geometric meaning of $(Z_1 . Z_2)$ is clear (at least over the complex numbers): take a generic topological perturbation and count transversal intersections with orientation; the interpretation of $\chi(X,L^* \otimes \mathcal{O}_{Z_2})$ is less so.

We somehow trade the torsion sheaf $\mathcal{O}_{Z_1}$
for the more "global" object $L^*$ (with $Supp(L^i)=X$). We certainly need some global information about $X$ to compute the intersection, as the generic perturbations take place in $X$.

Note that if $Z_1 \sim Z_1'$ is an algebraic generic perturbation, such that $Z_1' \cap Z_2$ is transversal (and zero-dimensional), then $\chi(Z_1' \otimes Z_2)=h^0(Z_1' \otimes Z_2)=(Z_1 . Z_2)$. So there is a formal analogy between $L^*$ and a generic perturbation of $Z_1$.

Addendum: There is yet another link between deformations of $Z_1$ and resolutions of $\mathcal{O}_{Z_1}$: If one studies the local deformations of $Z_1$ inside $X$ one begins by deforming the surjection $L^{-1} \rightarrow \mathcal{O}_X$ to get a family of subschemes. To obtain a flat family one has to ensure that the relations do deform also, i.e. that there is a deformation of the presentation $L^{-2} \rightarrow L^{-1} \rightarrow \mathcal{O}_X$.
I do not know, in this case we can always extend this deformation to the whole complex $L^*$.
In any case, you can obtain deformations of $Z_1$ by deforming the maps $d_i$ keeping the relation $d^2 =0$ and the cohomology sheaves $H^{-i}(L^*), i>0$ fixed.

I'm not a big expert, but one thing it obviously does is constructing the sheaf M from the locally free bundles (locally free = projective). For example, consider a skyscraper sheaf on a line. It's not a bundle, but it can be represented by just two bundles and a map multiplication by x.

So, formally, you're asking for a construction by locally free bundles and morphisms between them of your given complex in the derived category of coherent sheaves. How would you do that? Well, you find a bundle which concides with your sheaf on an open subset, then you extend it from that subset in a special way, I believe, by getting a perverse sheaf. Then you reduce your problem to the manifold of dimension one less, and so on.

So what you have is a proof of some theorem of the form "every element of derived category can be made out of perverse sheaves". I think we actually have the Riemann-Hilbert correspondence here.

I'll be grateful if some more knowledgeable person could fill in the omissions.

I think this important question deserves one more answer, even after all the excellent ones already given.

One can think of a free resolution as "approximating the module by the ring". This tautology has some corollaries:

If you want the resolutions to reveal geometric properties, you need the ring to be nice to start with. Examples of this have been explained in all the other answers. In fact, you get the most information if your ring is "smooth" (regular local or polynomial rings).

Flipping the argument, this means that "good module + bad ring = bad resolution". For example, let $M$ be a field $k$, viewed as a module over the ring $R =k[x,y]/(x^2,xy)$. Then a resolution of $M$ is always bad, the ranks of the free modules will increase exponentially, and no nice information about $M$ can be learned from those numbers. In fact, the information now flows the other way, it tells you how bad $R$ is. For example, it follows that $R$ can not be a complete intersection, because if it was, a result by Eisenbud implies that the Betti numbers would have had polynomial growth.